# Domains?

• Jan 9th 2007, 03:04 PM
leviathanwave
Domains?
Hi, I just did these two problems and the domain was the same for both problems, am I overlooking something?

h(x) = 4x/Square Root(x^2 - 16)
and
j(x) = Square Root[4x/(x^2 - 16)]

Maybe she's trying to trick me?
• Jan 9th 2007, 03:10 PM
leviathanwave
Oops!
One more quick thing. I didn't quite know how to approach this one:

The function y=k(x) is a polynomial with x-intercepts 1,2,5 and 8. The solution set to k(x) is (1, 2)U(5,8).

Use this information to find the domain of y= Square Root[k(x)]
• Jan 9th 2007, 04:44 PM
ThePerfectHacker
Quote:

Originally Posted by leviathanwave
Hi, I just did these two problems and the domain was the same for both problems, am I overlooking something?

h(x) = 4x/Square Root(x^2 - 16)
and

The denominator cannot be zero. And the radical must be positive. Thus,
$x^2-16>0$
$x^2>16$
$x>4 \mbox{ or }x<-4$
Quote:

j(x) = Square Root[4x/(x^2 - 16)]
Over here we need term inside the radical to be positive.
First $x^2-16\not = 0$ thus, $x\not = \pm 4$.
And,
$\frac{4x}{x^2-16}>0$
Both positive or both negative

1) $4x\geq 0 \to x\geq 0$
And,
$x^2-16 >0$ thus, $x>4 \mbox{ or }x<-4$
Combination of those two leads us to,
$x>4$

2) $4x\leq 0 \to x\leq 0$
$x^2-16 >0$ thus, $x>4 \mbox{ or }x<-4$.
Combination of those two leads us to,
$x<-4$.

Even though they are different functions they have the same solution set.
• Jan 9th 2007, 06:55 PM
topsquark
Quote:

Originally Posted by leviathanwave
One more quick thing. I didn't quite know how to approach this one:

The function y=k(x) is a polynomial with x-intercepts 1,2,5 and 8. The solution set to k(x) is (1, 2)U(5,8).

Use this information to find the domain of y= Square Root[k(x)]

Typically if you want a polynomial with specific x-intercepts you can simply write them out term by term:

If you have the intercepts 1, 2, 5, and 8 a polynomial that fits this is:
$f(x) = (x - 1)(x - 2)(x - 5)(x - 8)$

Now we want the function $y = \sqrt{k(x)}$ to have the domain $(1, 2) \cup (5, 8)$ which means that this is the part of the (real) x-axis for which k(x) is positive. So let's graph f(x). (See the attachment below.)

Note that in the graph the set $(1, 2) \cup (5, 8)$ is the region where f(x) is negative. So we need to "flip the function over" to make this the region where the function is positive. This is easy. Just define k(x) to be the negative of f(x):
$k(x) = -f(x) = -(x - 1)(x - 2)(x - 5)(x - 8)$.

Then $y = \sqrt{k(x)}$ has the correct domain.

Note: Technically the domain is $[1, 2] \cup [5, 8]$. I couldn't think of a way to define a finite polynomial that excludes the end-points of the sets. (Unless I'm missing the obvious.) You might wish to point this out to your teacher.

-Dan